Laboratory Report 1
What purpose do the resistors have in Step 1 and Step 2 of the diagram?
In steps 1 and 2 of the experiment our goal is to light up the LEDs or Light Emitting Diode. Now, to make an LED work, we need a voltage supply and a resistor. We need the voltage supply because the LED will only emit light when current runs through it. So what then is the use of the resistors?
LEDs are fun to use in experiments because they never burn out unless they exceed their current limit. That's where resistors come in. Resistors have a predetermined resistance so that it can control the amount of current that will flow through a component at a given time. If you try to use an LED without a resistor, most probably you'll end up burning it out after a few minutes or if a great amount current passes through it. The LED has a very little resistance so large amounts of current will try to flow through it unless you make use of a resistor.
A current running through from 0.02 Amps to 0.04 is of fairly good range, low enough not to burn out the LED. Using the Ohm's Law, we can know if the resistance is high enough to keep the current down. Ohm's law states that current is equal to the voltage drop / resistance (I = V/R). In our experiment, we use the fact that when the LED is on, there is a 1.2 voltage drop across it. This means, if the positive leg connects to +5 volts, the negative leg will be at 3.8 volts. Knowing the voltage on both sides of the resistor, we calculate the current:
Current = (5 volts -1.2 volts) / 470 W = 0.008 Amperes = 8 mA
The current is low enough not to burn out the LED!
How do you recognize 1's and 0's using the circuits you built in Step 3?
Give a brief explanation.
In order to recognize which is "1" and "0", we connect the end of the probe supposedly connected to 3 of the 555 to the ground. The light that turns on is equivalent to "0". After this, try to connect the same end of the probe to the +5 volts. The light that turns on is the "1".
The probe is connected between the LEDs in setup 3. When the probe is connected in +5, the LED that is nearer the +5 will not light up. Rather, it will be the LED farther from the +5 which will light up, because the connection of the LED nearer the +5 will become open and the only closed connection remaining at this time is the LED that is connected farther from the +5 (the LED that is nearer the ground). The same principle applies also when the probe is connected to the ground. But instead, the situation of the LEDs will reverse.
What is the purpose of the capacitor and resistor to control the
speed of the change from "1" and "0"? Give a brief explanation.
A capacitor is a passive electronic component that stores energy. Meanwhile, a resistor limits or regulates the flow of electrical current in an electronic circuit. The capacitors and the resistors have timing functions in changing the "1"s and "0"s. A low resistance results in a higher current flow and the capacitor will charge faster. Consequently, a high resistance results in a lower current flow, therefore it will take a longer time to charge. This means that if the capacitor takes a longer time to charge, the speed of the change form "1" to "0" will consequently be longer. If the capacitor takes a shorter time to charge, the change from "1" to "0" will also be shorter.
Thus, the speed of change will ultimately depend on the amount of resistance. In the experiment, we used a device known as a potentiometer, which we placed in parallel with the three resistors connected in series. It is basically a manually adjustable electrical resistor. With the turn of the knob, we can change the amount of resistance so that we control the speed of current flow. And this means, we can also control the speed of the change from "1"s to "0"s.
How does the 555 run an oscillator? Give a brief explanation.
Pin 2 and pin 6 of the 555 are connected so that they would know which guy would tell pin 3 to go either high or low. When the voltage is 2/3 or higher of the original, the pin 6 of the 555 will sense it and tell pin 3 to go low. It is like putting the probe on +5 Vcc and thus, lighting one of the LEDs at the same time discharging the capacitor. On the other hand, when we have 1/3 or lower of the original voltage, pin 2 will sense it and tell pin 3 to go high. At this time it is like putting the probe on ground and thus, lighting the other LED. Also, at this time, the capacitor is charging. This explains the alternating charging and discharging of the capacitor and the blinking of the lights alternately.
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